Abstract
The symmedian point of a triangle is known to give rise to two circles, ob- tained by drawing respectively parallels and antiparallels to the sides of the triangle through the symmedian point. In this note we will explore a third circle with a similar construction - discovered by Jean-Pierre Ehrmann [3]. It is ob- tained by drawing circles through the symmedian point and two vertices of the triangle, and intersecting these circles with the triangle’s sides. We prove the existence of this circle and identify its center and radius.
Ehrmann’s third Lemoine circle
Darij Grinberg written 2005; edited 2011
§1. The …rst two Lemoine circles
Let us remind the reader about some classical triangle geometry …rst. LetLbe the symmedian point of a triangle ABC. Then, the following two results1 are well-known ([2], Chapter 9):
A
B
C L
U
Fig. 1
Theorem 1 Let the parallels to the lines BC; BC; CA; CA; AB; AB through L meet the lines CA; AB; AB; BC; BC; CA at six points. These six points lie on one circle,
1For the de…nition of aTucker circle, see below.
the so-called …rst Lemoine circleof triangle ABC; this circle is a Tucker circle, and its center is the midpoint of the segment U L, where U is the circumcenter of triangle ABC. (See Fig. 1.)
[The somewhat uncommon formulation "Let the parallels to the lines BC; BC; CA;
CA; AB; AB through L meet the lines CA; AB; AB; BC; BC; CA at six points"
means the following: Take the point where the parallel to BC through L meets CA, the point where the parallel to BC throughL meetsAB, the point where the parallel to CA through L meetsAB, the point where the parallel toCA through L meets BC, the point where the parallel to AB through L meets BC; and the point where the parallel to AB through L meets CA.]
A
B
C L
Fig. 2
Furthermore (see [2] for this as well):
Theorem 2 Let the antiparallels to the lines BC; BC; CA; CA; AB; AB through L meet the lines CA; AB; AB; BC; BC; CA at six points. These six points lie on one circle, the so-called second Lemoine circle (also known as the cosine circle) of triangle ABC; this circle is a Tucker circle, and its center is L. (See Fig. 2.)
We have been using the notion of a Tucker circle here. This can be de…ned as follows:
Theorem 3 Let ABC be a triangle. Let Qa and Ra be two points on the line BC.
Let Rb and Pb be two points on the line CA. Let Pc and Qc be two points on the line AB. Assume that the following six conditions hold: The lines QbRc, RcPa, PbQa are parallel to the lines BC, CA, AB, respectively; the lines PbPc, QcQa, RaRb are antiparallel to the sidelines BC, CA, AB of triangle ABC, respectively. (Actually, requiring …ve of these conditions is enough, since any …ve of them imply the sixth one, as one can show.) Then, the points Qa, Ra, Rb, Pb, Pc and Qc lie on one circle. Such circles are called Tucker circlesof triangle ABC. The center of each such circle lies on the line U L, where U is the circumcenter and L the symmedian point of triangle ABC. Notable Tucker circles are the circumcircle of triangleABC, its …rst and second Lemoine circles (and the third one we will de…ne below), and its Taylor circle.
§2. The third Lemoine circle
Far less known than these two results is the existence of a third member can be added to this family of Tucker circles related to the symmedian point L. As far as I know, it has been …rst discovered by Jean-Pierre Ehrmann in 2002 [3]:
Theorem 4 Let the circumcircle of triangle BLC meet the lines CA and AB at the points Ab and Ac(apart from C andB). Let the circumcircle of triangleCLAmeet the lines AB and BC at the point Bc and Ba (apart from A and C). Let the circumcircle of triangle ALB meet the lines BC and CA at the points Ca and Cb (apart from B and A). Then, the six points Ab; Ac; Bc; Ba; Ca; Cb lie on one circle. This circle is a Tucker circle, and its midpoint M lies on the line U L and satis…es LM = 12 LU (where the segments are directed). The radius of this circle is 12p
9r12+r2; where r is the circumradius and r1 is the radius of the second Lemoine circle of triangle ABC.
We propose to denote the circle through the points Ab; Ac; Bc; Ba; Ca; Cb as the third Lemoine circleof triangle ABC. (See Fig. 3.)
A
B
C
L U
A
bA
cB
aB
cC
bC
aM
Fig. 3
The rest of this note will be about proving this theorem. First we will give a complete proof of Theorem 4 in §3-§5; this proof will use four auxiliary facts (Theorems 5, 6, 7 and 8). Then, in §6 and §7, we will give a new argument to show the part of Theorem 4 which claims that the six points Ab; Ac; Bc; Ba; Ca; Cb lie on one circle;
this argument will not give us any information about the center of this circle (so that it doesn’t extend to a complete second proof of Theorem 4, apparently), but it has the advantage of showing a converse to Theorem 4 (which we formulate as Theorem 10 in the …nal paragraph §8).
§3. A lemma
In triangle geometry, most nontrivial proofs begin by deducing further (and easier) properties of the con…guration. These properties are then used as lemmas (and even if they don’t turn out directly useful, they are often interesting for themselves). In the case of Theorem 4, the following result plays the role of such a lemma:
Theorem 5 The point L is the centroid of each of the three trianglesAAbAc; BaBBc; CaCbC: (See Fig. 4.)
A
B
C L
A
bA
cB
aB
cC
bC
aFig. 4
Actually this result isn’t as much about symmedian points and centroids, as it generalizes to arbitrary isogonal conjugates:
Theorem 6 Let P andQ be two points isogonally conjugate to each other with respect to triangle ABC. Let the circumcircle of triangle CP A meet the lines AB and BC at the points Bc andBa (apart fromA andC). Then, the triangles BaBBc and ABC are oppositely similar, and the points P and Q are corresponding points in the triangles BaBBc and ABC: (See Fig. 5.)
Remark. Two points P1 and P2 are said to be corresponding points in two similar triangles 1 and 2 if the similitude transformation that maps triangle 1 to triangle
2 maps the point P1 to the pointP2.
A
B
C P
Q B
cB
aw
αw
γFig. 5
Proof of Theorem 6. We will use directed angles modulo 180 . A very readable introduction into this kind of angles can be found in [1]. A list of their important properties has also been given in [4].
The point Qis the isogonal conjugate of the pointP with respect to triangleABC;
thus,]P AB =]CAQ.
Since C; A; Bc; Ba are concyclic points, we have ]CBaBc = ]CABc; so that ]BBaBc = ]BAC: Furthermore, ]BcBBa = ]CBA: Thus, the triangles BaBBc and ABC are oppositely similar (having two pairs of oppositely equal angles).
By the chordal angle theorem,]P BaBc=]P ABc=]P AB =]CAQ= ]QAC.
Similarly, ]P BcPa = ]QCA. These two equations show that the triangles BaP Bc and AQC are oppositely similar. Combining this with the opposite similarity of tri- angles BaBBc and ABC, we obtain that the quadrilaterals BaBBcP and ABCQ are oppositely similar. Hence, P and Q are corresponding points in the triangles BaBBc and ABC. (See Fig. 6.) Theorem 6 is thus proven.
A
B
C P
P' B
cB
aFig. 6
Proof of Theorem 5. Now return to the con…guration of Theorem 4. To prove Theorem 5, we apply Theorem 6 to the case whenP is the symmedian point of triangle ABC; the isogonal conjugate Q of P is, in this case, the centroid of triangle ABC.
Now, Theorem 6 says that the pointsP andQare corresponding points in the triangles BaBBc and ABC: Since Q is the centroid of triangle ABC, this means thatP is the centroid of triangle BaBBc: But P = L; thus, we have shown that L is the centroid of triangle BaBBc. Similarly, L is the centroid of triangles AAbAc and CaCbC; and Theorem 5 follows.
§4. Antiparallels
Theorem 5 was the …rst piece of our jigsaw. Next we are going to chase some angles.
Since the points B; C; Ab; Ac are concyclic, we have ]CAbAc = ]CBAc; so that ]AAbAc = ]ABC: Thus, the line AbAc is antiparallel to BC in triangle ABC.
Similarly, the lines BcBa and CaCb are antiparallel to CA and AB. We have thus shown:
Theorem 7 In the con…guration of Theorem 4, the lines AbAc; BcBa; CaCb are an- tiparallel to BC; CA; AB in triangleABC:
Now let Xb; Xc; Yc; Ya; Za; Zb be the points where the antiparallels to the lines BC; BC; CA; CA; AB; AB through L meet the lines CA; AB; AB; BC; BC; CA:
According to Theorem 2, these points Xb; Xc; Yc; Ya; Za; Zb lie on one circle around
L; however, to keep this note self-contained, we do not want to depend on Theorem 2 here, but rather prove the necessary facts on our own (Fig. 7):
Partial proof of Theorem 2.
A
B
C L
Y
aZ
aY
cX
cZ
bX
bFig. 7
Since symmedians bisect antiparallels, and since the symmedian pointLof triangle ABClies on all three symmedians, the pointLmust bisect the three antiparallelsXbXc; YcYa; ZaZb: This means that LXb = LXc; LYc = LYa and LZa = LZb: Furthermore, ]AXcXb = ]ACB (since XbXc is antiparallel to BC), thus ]YcXcL = ]ACB;
similarly, ]XcYcL = ]BCA; thus ]LYcXc = ]XcYcL = ]BCA = ]ACB = ]YcXcL: Hence, triangleXcLYc is isosceles, so thatLXc=LYc: Similarly, LZb =LXb and LYa =LZa: Hence,
LXb =LXc =LYc =LYa =LZa =LZb:
This shows that the points Xb; Xc; Yc; Ya; Za; Zb lie on one circle around L. This circle is the so-called second Lemoine circle of triangle ABC. Its radius isr1 =LXb = LXc=LYc=LYa=LZa=LZb:
We thus have incidentally proven most of Theorem 2 (to complete the proof of Theorem 2, we would only have to show that this circle is a Tucker circle, which is easy); but we have also made headway to the proof of Theorem 4.
A
B
C
L U B
aB
cM
B
mY
cY
aFig. 8
Proof of Theorem 9, part 1. Now consider Fig. 8. Let Bm be the midpoint of the segment BcBa: According to Theorem 5, the point L is the centroid of triangle BaBBc; thus, it lies on the median BBm and divides it in the ratio 2 : 1. Hence, BL:LBm = 2 (with directed segments).
Let M be the point on the line U L such that LM = 12 LU (with directed segments); then, LU = 2 LM;so that U L= LU = 2 LM; and thus U L:LM = 2 =BL :LBm:By the converse of Thales’theorem, this yields BmM kBU:
The linesYcYaandBcBaare both antiparallel toCA, and thus parallel to each other.
Hence, Thales’ theorem yields BmBc : LYc = BBm : BL: But since BL : LBm = 2, we have BL = 2 LBm; so that LBm = 12 BL and therefore BBm = BL+LBm = BL + 12 BL = 32 BL and BBm : BL = 32: Consequently, BmBc : LYc = 32 and BmBc= 32 LYc= 32r1:
It is a known fact that every line antiparallel to the side CA of triangle ABC is perpendicular to the line BU (where, as we remind,U is the circumcenter of triangle ABC). Thus, the line BcBa (being antiparallel to CA) is perpendicular to the line BU. Since BmM k BU, this yields BcBa ? BmM: Furthermore, BmM k BU yields BU : BmM = BL : LBm (by Thales), so that BU : BmM = 2 and BU = 2 BmM;
and thus BmM = 12 BU = 12r; where r is the circumradius of triangle ABC.
Now, Pythagoras’theorem in the right-angled triangle M BmBc yields M Bc=p
BmBc2+BmM2 = s
3 2r1
2
+ 1 2r
2
= r9
4r21+ 1
4r2 = 1 2
q
9r21+r2:
Similarly, we obtain the same value 12p
9r12+r2 for each of the lengths M Ba; M Ca;
M Cb; M Ab and M Ac. Hence, the points Ab; Ac; Bc; Ba; Ca; Cb all lie on the circle with centerM and radius 12p
9r21+r2: The point M, in turn, lies on the line U L and satis…es LM = 12 LU:
This already proves most of Theorem 4. The only part that has yet to be shown is that this circle is a Tucker circle. We will do this next.
§5. Parallels
A
B
C L
A
bA
cB
aB
cC
bC
aFig. 9
Since the points Ab; Ba; Ca; Cb are concyclic, we have ]CaBaAb = ]CaCbAb; thus ]CBaAb =]CaCbC: But since CaCb is antiparallel to AB, we have ]CCbCa = ]CBA; so that ]CaCbC = ]CCbCa = ]CBA; thus ]CBaAb = ]CBA; conse- quently,AbBa kAB: Similarly, BcCb kBC and CaAckCA. Altogether, we have thus seen:
Theorem 8 The lines BcCb; CaAc; AbBa are parallel to BC; CA; AB: (See Fig. 9.) Proof of Theorem 4, part 2. If we now combine Theorem 7 and Theorem 8, we conclude that the sides of the hexagonAbAcCaCbBcBa are alternately antiparallel and
the circle passing through its vertices Ab; Ac; Bc; Ba; Ca; Cb is a Tucker circle. This concludes the proof of Theorem 4.
One remark: It is known that the radius r1 of the second Lemoine circle ABC is rtan!; where ! is the Brocard angle of triangle ABC. 2 Thus, the radius of the third Lemoine circle is
1 2
q
9r21+r2 = 1 2
q
9 (rtan!)2+r2 = 1 2
p9r2tan2!+r2 = r 2
p9 tan2!+ 1:
§6. A di¤erent approach: a lemma about four points
At this point, we are done with our job: Theorem 4 is proven. However, our proof depended on the construction of a number of auxiliary points (not onlyBm, but also the six points Xb; Xc; Yc; Ya; Za; Zb). One might wonder whether there isn’t also a (possibly more complicated, but) more straightforward approach to proving the concyclicity of the points Ab; Ac; Bc; Ba; Ca; Cb without auxiliary constructions. We will show such an approach now. It will not yield the complete Theorem 4, but on the upside, it helps proving a kind of converse.
2For a quick proof of this fact, letX,Y, Z denote the feet of the perpendiculars from the point L to the sides BC, CA, AB of triangle ABC. Then, the radius r1 of the second Lemoine circle is r1 = LYa. Working without directed angles now, we see that LYa = LX
sinA (from the right-angled triangle LXYa), so that r1 = LYa = LX
sinA. Since sinA = a
2r by the extended law of sines, this rewrites as r1 = LX
a 2r
= LX 2r
a . This rewrites as a2
2rr1 =a LX. Similarly, b2
2rr1 =b LY and c2
2rr1=c LZ. Adding these three equations together, we obtain a2
2rr1+b2 2rr1+c2
2rr1=a LX+b LY +c LZ:
On the other hand, letS denote the area of triangleABC. But the area of triangleBLC is 1 2a LX (since a is a sidelength of triangle BLC, andLX is the corresponding altitude), and similarly the areas of trianglesCLAandALB are 1
2b LY and 1
2c LZ. Thus, 1
2a LX+1
2b LY +1
2c LZ= (area of triangleBLC) + (area of triangleCLA) + (area of triangle ALB)
= (area of triangleABC) =S;
so thata LX+b LY +c LZ= 2S. The equation a2 2rr1+ b2
2rr1+ c2
2rr1=a LX+b LY +c LZ thus becomes a2
2rr1+ b2 2rr1+ c2
2rr1= 2S, so that
r1= 2S a2
2rr1+b2 2rr1+c2
2rr1
= 4rS
a2+b2+c2 =r a2+b2+c2
| 4S{z }
=cot!
=r cot!=rtan!;
qed.
A B
C
D
Fig.
10
Let us use directed areas and powers of points with respect to circles. Our main vehicle is the following fact:
Theorem 9 Let A; B; C; D be four points. Let pA; pB; pC; pD denote the powers of the pointsA; B; C; D with respect to the circumcircles of trianglesBCD; CDA; DAB;
ABC. Furthermore, we denote by [P1P2P3] the directed area of any triangle P1P2P3: Then,
pA [BCD] =pB [CDA] = pC [DAB] =pD [ABC]: (1) (See Fig. 10.)
Proof of Theorem 9.. In the following, all angles are directed angles modulo 180 , and all segments and areas are directed. Let the circumcircle of triangle BCD meet the line AC at a point A0 (apart from C). Let the circumcircle of triangle CDA meet the line BD at a point B0 (apart from D). Let the lines AC and BD intersect at P. Since the points C; D; A; B0 are concyclic, we have ]DB0A=]DCA;
thus ]P B0A = ]DCA: But since the points C; D; B; A0 are concyclic, we have ]DBA0 = ]DCA0; so that ]P BA0 = ]DCA: Therefore, ]P B0A = ]P BA0; which leads toB0AkBA0. Hence, by the Thales theorem,
A0A
BB0 = P A0 P B:
A B
C
D P
B'
A'
Fig. 11 The power pA of the point A with respect to the circumcircle of triangle BCD is AA0 AC; similarly, pB =BB0 BD: Thus,
pA [BCD]
pB [CDA] = AA0 AC [BCD]
BB0 BD [CDA] = A0A AC [BCD]
BB0 BD [CDA] = A0A BB0
AC BD
[BCD]
[CDA]
= P A0 P B
AC BD
[BCD]
[CDA]:
But it is a known fact that wheneverA,B,C are three collinear points andP is a point not collinear withA,B,C, then we have AB
AC = [APB]
[APC]. This fact yields CP
CA = [CDP]
[CDA] and DP
DB = [DCP] [DCB]; so that
CP CA : DP
DB = [CDP]
[CDA] : [DCP]
[DCB] = [DCB] [CDP]
[DCP] [CDA] = [BCD] [CDP]
[CDP] [CDA] = [BCD]
[CDA];
and thus
pA [BCD]
pB [CDA] = P A0 P B
AC BD
[BCD]
[CDA] = P A0 P B
AC BD
CP CA : DP
DB
= P A0 P B
AC BD
P C AC : P D
BD = P A0 P B
AC BD
P C AC
BD
P D = P A0 P C P B P D: But the intersecting chord theorem yields P A0 P C =P B P D; so that
pA [BCD]
pB [CDA] = 1;
in other words, pA [BCD] =pB [CDA]: Similarly, pB [CDA] =pC [DAB]; so that pB [CDA] = pC [DAB]; and pC [DAB] = pD [ABC]: Combining these equalities, we get (1). This proves Theorem 9.
§7. Application to the triangle
Now the promised alternative proof of the fact that the points Ab; Ac; Bc; Ba; Ca; Cb are concyclic:
Proof. The identity (1) can be rewritten as
pA [BDC] =pB [CDA] =pC [ADB] =pD [ABC]
(since[BCD] = [BDC]and [DAB] = [ADB]). Applying this equation to the case when D is the symmedian point L of triangleABC, we get
pA [BLC] =pB [CLA] =pC [ALB]
(we have dropped the third equality sign since we don’t need it), where pA; pB; pC are the powers of the points A; B; C with respect to the circumcircles of triangles BLC; CLA; ALB. But we have pA = AC AAb and pB = BC BBa (Fig. 4); thus, pA [BLC] =pB [CLA] becomes
AC AAb [BLC] =BC BBa [CLA]; so that
AAb
BBa = BC AC
[CLA]
[BLC]: (2)
A
B
C L
Y
X Z
Fig. 12 Now letX; Y and Z be the feet of the perpendiculars from the symmedian pointL onto the linesBC; CA; AB. It is a known fact that the distances from the symmedian point of a triangle to its sides are proportional to these sides; thus, LX : LY :LZ = BC :CA:AB: 3. But since the area of a triangle equals 1
2 (one of its sidelengths) (corresponding altitude), we have [BLC] = 12 BC LX; [CLA] = 12 CA LY and [ALB] = 12 AB LZ:Thus, (2) becomes
AAb
BBa = BC AC
1
2 CA LY
1
2 BC LX = BC AC
CA BC
LY
LX = BC AC
CA BC
CA BC
= BC AC
CA2
BC2 = BC AC
AC2
BC2 = AC BC:
By the converse of Thales’ theorem, this yields AbBa k AB: Similarly, BcCb k BC and CaAc k CA; this proves Theorem 8. The proof of Theorem 7 can be done in the same way as §4 (it was too trivial to have any reasonable alternative). Combined, this yields that the sides of the hexagonAbAcCaCbBcBaAb are alternately antiparallel and parallel to the sides of triangleABC. Consequently, this hexagon is a Tucker hexagon, and since every Tucker hexagon is known to be cyclic, we thus conclude that the points Ab; Ac; Bc; Ba; Ca; Cb lie on one circle. This way we have reproven a part of Theorem 4.
Here is an alternative way to show that the points Ab; Ac; Bc; Ba; Ca; Cb lie on one circle, without using the theory of Tucker hexagons:
Proof. (See Fig. 9.) Since CaCb is antiparallel to AB, we have ]CCbCa =
3Actually, there is no need to refer to this known fact here, because we have almost completely proven it above. Indeed, while proving thatr1=rtan!, we showed thatr1= LX 2r
a , so thatLX= r1
2ra= r1
2rBC. Similarly,LY = r1
2rCAandLZ= r1
2rAB, so thatLX:LY :LZ =BC:CA:AB.
]CBA; and thus
]CbCaAc=](CaCb; CaAc) =](CaCb; CA) (since CaAckCA)
=]CaCbC= ]CCbCa =]CBA:
Since AbAc is antiparallel to BC, we have ]AAbAc = ]ABC; so that ]CbAbAc=]AAbAc = ]ABC =]CBA:
Therefore, ]CbAbAc = ]CbCaAc; so that the points Ca; Cb; Ab; Ac lie on one circle.
The point Ba also lies on this circle, since
]CaBaAb =](BC; AbBa) =](BC; AB) (sinceAbBa kAB)
=]CBA= ]CCbCa (since ]CCbCa= ]CBA was shown above)
=]CaCbAb:
Similarly, the pointBc lies on this circle as well. This shows that all six pointsAb; Ac; Bc; Ba; Ca; Cb lie on one circle.
This argument did never use anything but the facts that the linesAbAc; BcBa; CaCb are antiparallel to BC; CA; AB and that the lines BcCb; CaAc; AbBa are parallel to BC; CA; AB. It can therefore be used as a general argument why Tucker hexagons are cyclic.
§8. A converse
The alternative proof in §7 allows us to show a converse of Theorem 4:
Theorem 10 Let P be a point in the plane of a triangle ABC but not on its circum- circle. Let the circumcircle of triangle BP C meet the lines CA and AB at the points Ab and Ac (apart from C andB). Let the circumcircle of triangleCP A meet the lines AB and BC at the point Bc and Ba (apart from A and C). Let the circumcircle of triangle AP B meet the lines BC and CA at the points Ca and Cb (apart from B and A). If the six points Ab; Ac; Bc; Ba; Ca; Cb lie on one circle, then P is the symmedian point of triangle ABC.
We will not give a complete proof of this theorem here, but we only sketch its path:
First, it is easy to see that the lines AbAc; BcBa; CaCb are antiparallel to BC; CA;
AB. Now, we can reverse the argument from §7 to show that the lines BcCb; CaAc; AbBa are parallel toBC; CA; AB, and use this to conclude that the distances fromP to the sidelines BC, CA,AB of triangleABC are proportional to the lengths of BC;
CA; AB. But this implies that P is the symmedian point of triangleABC:
References
[1] R. A. Johnson, Directed Angles in Elementary Geometry, American Mathemat- ical Monthly vol. 24 (1917), #3, pp. 101-105.
[2] Ross Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, USA 1995.
[3] Jean-Pierre Ehrmann, Hyacinthos message #6098.
http://tech.groups.yahoo.com/group/Hyacinthos/message/6098
[4] Darij Grinberg, The Neuberg-Mineur circle, Mathematical Re‡ections 3/2007.
